Chapter 1

Introduction

Answers to Even Numbered Conceptual Questions
2. Atomic clocks are based on the electromagnetic waves that atoms emit. Also, pulsars are
highly regular astronomical clocks.
4. (a) ~
-1
0.5 lb 0.25 kg or ~10 kg
(b)
0
~4 lb 2 kg or ~10 kg
(c) ~
3
4000 lb 2000 kg or ~10 kg
6. Let us assume the atoms are solid spheres of diameter 10 . Then, the volume of each
atom is of the order of 10 . (More precisely, volume =
10
m
-30 3
m
3
4 1
=
3 6
r
3
d .) Therefore,
since 1 , the number of atoms in the solid is on the order of =
3 -6
cm 10 m
3

6
24
30
10
= 10
10

atoms. A more precise calculation would require knowledge of the density of the solid
and the mass of each atom. However, our estimate agrees with the more precise
calculation to within a factor of 10.
8. Realistically, the only lengths you might be able to verify are the length of a football field
and the length of a housefly. The only time intervals subject to verification would be the
length of a day and the time between normal heartbeats.
10. On the average, a typical person drives about 10 000 miles per year or about 30 miles per
day. (A reasonable estimate would be in the range of 5 to 50 miles per day.)
12. In general, the manufacturers are using significant figures correctly. The length and width
of the aluminum foil both include three significant figures and the cited value of their
product (the area) correctly includes three significant figures in both the metric and
English versions. The manufacturer of the tape also gives three significant figures in the
dimensions of the tape (metric version) and retains three significant figures in their
product (area). However, the width of the tape is given in English units as 1 2 , which
gives no indication of the accuracy of the measurement. To be consistent with the three
significant figure accuracy cited in the metric width dimension, the width should be given
as . 0.500
1
2 CHAPTER 1
Answers to Even Numbered Problems
2. (a)
2
L T (b) L
4. All three equations are dimensionally incorrect.
6. (a) (b)
-2
MLT
2
kg m s
8.
2 2
209 cm 4 cm
10. (a)
8
3.00 10 m s (b)
8
2.997 9 10 m s (c)
8
2.997 924 10 m s
12. (a) (b)
2
346 m 13 m
2

2 2
66.0 m 1.3 m
14. (a) (b)
9
2.96 10

2
6.876 10
16. (a) , ,
2
5.60 10 km
5
5.60 10 m
7
5.60 10 cm
(b) , , 0.4912 km 491.2 m
4
4.912 10 cm
(c) , 6.1 , 6.192 km
3
92 10 m
5
6.192 10 cm
(d) , , 2.499 km
3
2.499 10 m
5
2.499 10 cm
18. 9.2 nm s
20.
9
3 10 yr
22. ,
2 3
2.9 10 m
8 3
2.9 10 cm
24.
6 3
2.57 10 m
26. (a) = 1 mi h 1.609 km h
(b) 88 km h
(c) 16 km h
28. It would require about 47.6 yr to count the money. We advise against it.
30. of beef, ~ head of cattle (assumes 0.25 lb per burger and a net of 300 lb of
meat per head of cattle)
10
~10 lb
7
10
32. (assumes
7
~10 blades
2
1 16 in per blade)
34.
10
~10 cans yr ,
5
~10 ton yr (Assumes an average of 1 can per person each week, a
population of 250 million, and 0.5 oz of aluminum per can)
36. 2.2 m

Introduction 3
38. 8.1 cm
40. 2.33 m
42. (a) 1.50 m (b) 2.60 m
44. 8.60 m
46. 70.0 m
48. (a)
3
0.677 g cm (b)
17 2
4.63 10 ft
50. (a) (b)
2
~10 kg
3
~10 kg
52. (a) (b)
2
~10 yr
4
~10 times
54. (a) 4 (b) 8
4 CHAPTER 1
Problem Solutions
1.1 (a) The units of volume, area and height are:

, [ ] , and =
3
[ ] L V =
2
L A = [ ]

We then observe that
L h
=
3
L L
2
L or = [ ]

Thus, the equation
[ ][ ] V A h
= V is dimensionally correct. Ah
(b)
( )
= = =
2 2
cylinder
h R h V R , where Ah =
2
A R

( ) = = A A
rectangular box
V wh w = h Ah , where = = A A w length width
1.2 (a) From x B , we find that =
2
t =
2
x
B
t
. Thus, B has units of

= =
2
[ ]
[ ]
x
t
[ ] B
2
L
T

(b) If ( ) = sin 2 ft x A , then ( ) = [ ] [sin 2 ] [ ] A x ft

But the sine of an angle is a dimensionless ratio.

Therefore, = = ] [ ] [ A x L
1.3 Substituting in dimensions, we have

= =
2
2
L
T T
L T
= T

Thus, the dimensions are consistent
1.4 In the equation = +
2 2
1 1
0 2 2
mv mv mgh ,
| |
= = =
|
\ .
2
2
2 2
0 2
L M
[ ] [ ] M
T T
mv mv
L

while
| |
(
= =
|

\ .
1
2
2
L M
M L
T T
mgh
L
. Thus, the equation is dimensionally incorrect
In , = +
2
0
v v at = =
0
L
[ ]
T
v v [ ] but
( )
| |
= =
|
\ .
2 2 2
2
L
[ ][ ] T L
T
at a t = [ ] . Hence, this equation is
d y incorrect imensionall

Introduction 5
In the equation , we see that =
2
ma v
| |
= = =
|
\ .
2
L M
[ ][ ] M
T T
ma m a
2
L
[ ] while
| |
= =
|
\ .
2
2
2
2
L L
[ ]
T T
v
Therefore, this equation is also dimensionally incorrect
)
2
2
a F
1.5
(
( )

| |
|
\ .
2
kg kg m
=
s m
G
, so cross-multiplying gives the units of G as

3
2
m
kg s

1.6 (a) Given that m, we have F ma . Therefore, the units of force are those of ma,
( )
= = =
2
[ ] [ ] M L T F ma = [ ][ ] m a
-2
MLT
(b) newton =

2
kg m
s

1.7 (a) has 78.9 0.2 3 significant figures
(b) 3.7

has
9
88 10 4 significant figures
(c) 2.4 has

6
6 10 3 significant figures
(d) 0 has

=
3
.0032 3.2 10 2 significant figures
1.8 . Multiplying out this product of binominals
gives


( ) ( ) = = (

A 21.3 0.2 cm 9.8 0.1 cm A w
( ) ( )
(

( ) ( ) + (

2
0.2 0.1 cm = 21.3 9.8 21.3 0.1 0.2 9.8 A
=


The first term gives the best value of the area. The second and third terms add together
to give the uncertainty and the fourth term is negligible in comparison to the other
terms. The area and its uncertainty are found to be

A
2 2
209 cm 4 cm
1.9 (a) The sum is rounded to 797 because 756 in the terms to be added has no positions
beyond the decimal.
(b)
( )
=
- 3
.0032 356.3 3.2 10 356.3=1.14016 0 must be rounded to 1.1 because

has only two significant figures.
-3
3.2 10
6 CHAPTER 1
(c) .620 5 must be rounded to 17.66 because 5.620 has only four significant
figures.
1.10 =
8
2.997 924 574 10 m s c
(a) Rounded to 3 significant figures: c =
8
3.00 10 m s
(b) Rounded to 5 significant figures: c =
8
2.997 9 10 m s
(c) Rounded to 7 significant figures: c =
8
2.997 924 10 m s
1.11 The distance around is + + + = 38.44 m 19.5 m 38.44 m 19.5 m 115.88 m, but this answer
must be rounded to 115.9 m .5 m because the distance 19 carries information to only
one place past the decimal.
1.12 (a) ( ) ( ) ( )( ) ( )
(
= = = +

2 2
2
10.5 m 0.2 m 10.5 m 2 10.5 m 0.2 m 0.2 m A r
=
2

giving A
2 2
346 m 13 m
(b) ( ) = = 2 2 10.5 m 0.2 m = C r 66.0 m 1.3 m
1.13 Adding the two lengths together, we get 228.76 cm. However, 135.3 cm has only one
decimal place. Therefore, only one decimal place accuracy is possible in the sum,
changing 228.76 cm to 228.8 cm .
1.14 (a)
( )( ) ( )
=
4 9 4
2.437 10 6.5211 10 5.37 10 2.9594 10 =
9

9
2.96 10

(b)
( )( ) ( )

=
2 2 4 6
3.14159 10 27.01 10 1234 10 6.8764 10 =

2
6.876 10

1.15
( )
| || |
= =
| |
\ . \ .
5 280 ft 1 fathom
250 000 mi
1.000 mi 6 ft
d
8
2 10 fathoms

The answer is limited to one significant figure because of the accuracy to which the
conversion from fathoms to feet is given.

Introduction 7
1.16 (a) ( )
2 5
1.609 km
348 mi 5.60 10 km 5.60 10 m 5.60 10 cm
1.000 mi
| |
= = = =
|
\ .
A
7

(b) ( )
4
1.609 km
1612 ft 0.491 2 km 491.2 m 4.912 10 cm
5 280 ft
h
| |
= = = =
|
\ .

(c) ( )
3 5
1.609 km
20 320 ft 6.192 km 6.192 10 m 6.192 10 cm
5 280 ft
| |
= =
|
\ .
h = =
(d) ( )
3 5
1.609 km
8 200 ft 2.499 km 2.499 10 m 2.499 10 cm
5 280 ft
| |
= = =
|
\ .
d =
In (a), the answer is limited to three significant figures because of the accuracy of the
original data value, 348 miles. In (b), (c), and (d), the answers are limited to four
significant figures because of the accuracy to which the kilometers-to-feet conversion
factor is given.
1.17 ( )( ) ( )
| |
= = =
|
\ .
A
2
4 2
1 m
100 ft 150 ft 1.50 10 ft
3.281 ft
A w =
3 2
1.39 10 m

1.18
9
2
1 day 1 in 1 h 2.54 cm 10 nm
9.2 nm s
32 day 24 h 3 600 s 1.00 in 10 cm
rate
| | | | | |
| | | |
= =
| | | | |
\ . \ .
\ . \ . \ .


This means that the proteins are assembled at a rate of many layers of atoms each
second!

1.19
( )
| |
= =
|
\ .
16
3.281 ft
Distance 4 10 m
1 m

17
1 10 ft
1.20
( )
| |
| || |
= =
| | |
\ . \ .
\ .
17
1 yr 1 day 1 h
Age of earth 1 10 s
365.242 days 24 h 3600 s

9
3 10 yr
1.21
| || || ||
=
| | |
\ .\ .\ .\
8
3
m 3600 s 1 km 1 mi
3.00 10
s 1 h 10 m 1.609 km
c
|
=
|
.

8
6.71 10 mi h
8 CHAPTER 1
1.22 ( )( )( )
( )

| |
=
|
\ .
| |
= = =
|
\ .
2 3
3
3
2 2 3 3

2.832 10 m
Volume of house 50.0 ft 26 ft 8.0 ft
1 ft
100 cm
2.9 10 m 2.9 10 m 2.9 10 cm
1 m
8 3

1.23 ( )
3 2
4 3

43 560 ft 1 m
Volume 25.0 acre ft 3.08 10 m
1 acre 3.281 ft
| |
| |
= =
| |
\ .
\ .

1.24 ( )( )
( )( ) ( )
( )

=
(
= =

| |
= =
|
\ .
7 2 3
2 3
7 6 3 3
3
1
Volume of pyramid area of base height
3
1
13.0 acres 43 560 ft acre 481 ft 9.08 10 ft
3
2.832 10 m
9.08 10 ft 2.57 10 m
1 ft

1.25 (Where L = length of one side of the cube.)

Thus,
= =
3
Volume of cube 1 quart L
( )
| |
|
\ .
3
1 gallon
1 quart
4 quarts
L
= L
| || |
= =
| |
\ . \ .
3
3
3.786 liter 1000 cm
946 cm
1 gallon 1 liter


and 9.82 cm
1.26 (a)
| || |
= =
| |
\ .\ .
mi mi 1.609 km
1 1
h h 1 mi
km
1.609
h

(b)
| |
| |
= =
|
\ .
\ .
max
1.609 km h mi mi
55 55
h h 1 mi h
v =
|
km
88
h

(c)
| |
| |
= =
|
\ .
\ .
max
1.609 km h mi mi mi
65 55 10
h h h 1 mi h
v =
|
km
16
h

1.27 (a)
( )( ) ( )
( )

| |
= =
|
\ .
| |
| |
= =
| |
\ .
\ .
3
3
3
3
2

3 3
3
1.0 10 kg
mass density volume 1.0 m
1.0 cm
kg 10 cm
1.0 10 1.0 m 1.0 10 kg
cm 1 m
3


Introduction 9
(b) As rough calculation, treat as if 100% water.

cell:
( )

| |
=
|
\ .
3
3 6
3
kg 4
ss density volume= 10 0.50 10 m
m 3
= ma

16
5.2 10 kg

kidney:
( )

| |
= =
|
\ .
3
3 2
3
kg 4
mass density volume 10 4.0 10 m
m 3
= 0.27 kg

fly:
( )( )
( ) ( )

= =
| |
= =
|
\ .
2
2
5 3 3 3
3
mass density volume density
kg
10 1.0 10 m 4.0 10 m 1.3 10 kg
m
r h
1.28 Since you have only 16 hours (57 600 s) per day, you can count only $57 600 per day. So,
the time it would take would be:


| |

=
|

\ .
9
4
1 yr 1.00 10 dollars
5.76 10 dollars/day 365 days
t 47.6 yr

Right now, you are at least 18 yr old, so you would be at least age 65 when you finished
counting the money. It would provide a nice retirement, but a very boring life until then.
We would advise against it .
1.29 ( )( )( ) =
| || ||
| |
=
| | |
\ .
\ .\ .\
=
3
number of balls needed number lost per hitter number hitters games
innings games 1 hitters
ball per hitter 10 9 81
4 inning game year
balls balls
1800 or ~10
year year
|
|
.


Assumptions are 1 ball lost for every four hitters, 10 hitters per inning, 9 innings per
game, and 81 games per season.
10 CHAPTER 1
1.30 ( )( ) = number of pounds number of burgers weight burger
(



)( ) = =
10 10
5 10 burgers 0.25 lb burger 1.25 10 lb , or
10
~10 lb

( ) ( ) = er of head of cattle weight needed weight per head
(
numb


) ( ) = =
10 7
1.255 10 lb 300 lb head 4.17 10 head , or
7
~10 head

Assumptions are 0.25 lb of meat per burger and a net of 300 lb of meat per head of cattle
1.31 A reasonable guess for the diameter of a tire might be 2.5 ft, with a circumference of
about 8 ft. Thus, the tire would make

( )( )( ) =
7
50 000 mi 5280 ft mi 1 rev 8 ft 3 10 rev , or
7
~10 rev
1.32 A blade of grass is ~1 4 inch wide, so we might expect each blade of grass to require at least

=
2
1 16 in 4.3 10 ft
4 2
. Since , the number of blades of grass to be
expected on a quarter-acre plot of land is about


=
2
1 acre 43 560 ft
( )( )

- 4 2
0.25 acre 43 560 ft
4.3 10 ft bl
r


=
7
2.5 10 blades, o
7
~10 blades
= =
2
acre
total area
area per blade ade
n
1.33 Consider a room that is 12 ft square with an 8.0 ft high ceiling. The volume of this room
is

( )( )( )
| |
= =
|
\ .
3
3
1 m
12 ft 12 ft 8.0 ft 33 m
3.281 ft
room
V

A ping pong ball has a radius of about 2.0 cm, so its volume is


( )


= = =
3
3 2
4 4
2.0 10 m 3.4 10 m
3 3
ball
V r
5 3


The number of balls that would easily fit into the room is therefore

= = =

3
5
-5 3
33 m
9.7 10
3.4 10 m
room
ball
V
n
V
or
6
~10

Introduction 11
1.34 Assume an average of 1 can per person each week and a population of 250 million.

( )( )
( )( )
| |
=
|
\ .
| |
=
|
\ .
=
8
10 10
number cans person
number cans year population weeks year
week
can person
1 2.5 10 people 52 weeks yr
week
1.3 10 cans yr , or ~10 cans yr


( )( ) =
| |
(
| || || |
=
| | | |
(
\ .\ .\ .

\ .
=
10
5 5
number of tons weight can number cans year
oz 1 lb 1 ton can
0.5 1.3 10
can 16 oz 2000 lb yr
2 10 ton yr , or ~10 ton yr


Assumes an average weight of 0.5 oz of aluminum per can.
1.35 The x coordinate is found as ( ) = = =
D
cos 2.5 m cos35 x r 2.0 m

and the y coordinate ( ) = = =
D
2.5 m sin35 sin y r 1.4 m
1.36 The x distance out to the fly is 2.0 m and the y distance up to the fly is 1.0 m. Thus, we
can use the Pythagorean theorem to find the distance from the origin to the fly as,

( ) ( ) = + +
2 2
2 2
= 2.0 m 1.0 m = d x y 2.2 m
1.37 The distance from the origin to the fly is r in polar coordinates, and this was found to be
2.2 m in Problem 36. The angle is the angle between r and the horizontal reference line
(the x axis in this case). Thus, the angle can be found as

= = =
1.0 m
tan 0.5
2.0 m
y
x
and ( )

= =
1
tan 0.50 27

The polar coordinates are = = 2 m and 27 2. r
12 CHAPTER 1
1.38 The x distance between the two points is 8.0 cm and the y distance between them is 1.0
cm. The distance between them is found from the Pythagorean theorem:

( ) ( ) = + = + =
2 2
2 2 2
8.0 cm 1.0 cm 65 cm = d x y 8.1 cm
1.39 (a) From the Pythagorean theorem, the unknown side is

( ) ( ) = =
2 2
2 2
9.00 m 6.00 m = b c a 6.71 m
b) tan = =
6.00 m
6.71 m
0.894 (c) = =
6.71 m
n
9.00 m
si 0.746
1.40 From the diagram, ( ) = cos 75.0 d L

Thus, ( ) ( ) ( ) = = 9.00 m cos = cos 75.0 75.0 d L 2.33 m







1.41 The circumference of the fountain is = 2 C r , so the radius is
d
75.0
L

=

9
.
0
0

m
55.0
h
r


= = =
15.0 m
2.39 m
2 2
C
( )
r

Thus, = =
2.39 m
h h
r
tan 55.0 which gives

( ) ( ) = = n 55.0 2 h .39 m ta 3.41 m
1.42 (a) =
side opposite
sin
hypotenuse
so, ( )( ) = = de opposite 3.00 m sin30.0 si 1.50 m
(b) = co so,
adjacent side
s
hypotenuse
( )( ) = = adjacent side 3.00 m cos30.0 2.60 m

Introduction 13
1.43 (a) The side opposite = 3.00 (b) The side adjacent to = 3.00
(c) = = co
4.00
s
5.00
0.800 (d) = =
4.0
si
0
n
5.00
0.800
(e) = = tan
4.00
3.00
1.33
1.44 Using the diagram at the right, the Pythagorean Theorem
yields

( ) ( ) = +
2 2
5.00 m 7.00 m c = 8.60 m
1.45 From the diagram given in Problem 1.44 above, it is seen that

= tan
5.00
7.00
and = 35.5
5.00 m
7.00 m
c
q
100 m
Tree
w
River
35.0
1.46 Using the sketch at the right:

= tan35.0 , or
100 m
w
(


) = 100 m tan35.0 w = 70.0 m
1.47 The c as can be seen by considering the ratio of the areas of the
two pizzas.


ustomer is incorrect
( )
( )

= = = =
2 2
2 2
9 in 81 9
36 4 6 in
large large
small small
A r
A r


If the small pizza costs $6.00, the cost of the large one should be ( )
| |
=
|
\ .
9
$6.00
4
$13.50
14 CHAPTER 1
1.48 (a) The volume of Saturn is
( )
= =
3
7 2 3
4 4
= 5.85 10 m 8.39 10 m
3 3
3 3
V r and the
density is


| | | |
| |
=
| | |
\ .
\ . \ .
3 26 3
23 3 2
5.68 10 kg 10 g 1 m
=
8.39 10 m 10 cm 1 kg
m
V
3
0.677 g cm .
(b) The surface area of Saturn is


( )

| |
= =
|
\ .
2
2
2 7
3.281 ft
4 4 5.85 10 m
1 m
A r =
17 2
4.63 10 ft

1.49 The term s has dimensions of L, a has dimensions of
2
LT
2 m n m
, and t has dimensions of T.
Therefore, the equation, s k has dimensions of



or

The powers of L and T must be the same on each side of the equation. Therefore,
=
m n
a t
( ) ( ) =
- 2
L LT T
m
n
=
1 0
L T L T
=
1
L L
m

and = 1 m

Likewise, equating terms in T, we see that = 2 n m 0 . Thus, = = 2 2 n m

, a dimensionless constant. Dimen min sional analysis cannot deter e the value of k
1.50 Assume the tub measures 1.3 m by 0.5 m by 0.3 m
(a) It then has a volume V and contains a mass of water


=
3
0.2 m
( )( )
= = =
3
10 kg 200 kg, or m V
3 3
m 0.2 m
2
~10 kg
(b) Pennies are now mostly zinc, but consider copper pennies filling 80% of the volume
of the tub. Their mass is


( )( )
= =
3 3 3
0.80 8.93 10 kg/m 0.2 m 1400 kg, or m
3
~10 kg

Introduction 15
1.51 The volume of oil equals

= =
-7

-10 3
3
9.00 10 kg
9.80 10 m
918 kg m
V

If the diameter of a molecule is d, then this volume must also equal


( ) ( )( s of slick area of oil slick) =
2
thicknes d r , where = 0.418 m r

Thus,
( )

= =
10 3
9
2
9.80 10 m
1.78 10 m, or
0.418 m
d
9
~10 m
1.52 (a) The amount paid per year would be


| || |

| |
= =
| | |
\ .
\ .\ .
4
10
365.25 days dollars 8.64 10 s dollars
amount 1000 3.16 10
s 1.00 day yr yr


Therefore, it would take

12
10
7.00 10 dollars
222 yr, or
3.16 10 dollars yr
2
~10 yr
(b) The circumference of the Earth at the equator is


( )
= = =
7 6
2 2 6.38 10 m 4.01 10 m C r

12
10 m


The length of one dollar bill is 0.155 m so that the length of seven trillion bills is
. Thus, the seven trillion dollars would encircle the earth


1.09

= =

12
4
7
1.09 10 m
2.71 10 , or
4.01 10 m
n
4
~10 times
1.53 The number of tuners is found by dividing the number of residents of the city by the
number of residents serviced by one tuner. We shall assume 1 tuner per 10,000 residents
and a population of 7.5 million. Thus,

6
2
4
7.5 10
number of tuners =7.5 10
1.0 10
3
~10 tuners
1.54 (a) For a sphere, =
2
4 A R . In this case, =
2 1
2 R R
Hence,
( )

= =
2 2
2 2
2 2
1 1
4
4
R R
A R
=
2
1
2
2
1 1
2R
A
R R
= 4
16 CHAPTER 1

(b) For a sphere, =
3
4
3
V R

Thus,
( )
( )
( )

= = =
3
3 3
2 1 2 2
3 3 3
1 1 1
4 3 2
4 3
R R R V
V R R R
=
1
8
1.55 (a) ( )
| ||

|
=
|
\ .\
4
365.2 days 8.64 10 s
r= 1 yr
1 yr 1 day
|
.
1 y
7
3.16 10 s
(b) Let us consider a segment of the surface of the moon which has an area of 1 m and
a depth of 1 . When filled with meteorites, each having a diameter 10 , the
number of meteorites along each edge of this box is


2
m m
-6
= =
6
-6
len of an edge 1 m
10
me ite diameter 10 m
n =
gth
teor
(


The total number of meteorites in the filled box is then


)
= = 10 =
3 6 18 3
10 N n
( )


At the rate of 1 meteorite per second, the time to fill the box is


| |
= =
|

\ .
18 18
7
1 y
10 s 3 10
3.156 10 s
t
10
yr, or s= 10
10
~10 yr